X(3026) = 10th STEVANOVIC
POINT

X(3026) lies on the incircle and these lines:
150,388 274,1682 1111,1357
1365,1565

Antipodal Pairs on Circles

In response to Stevanovic's findings (X(3000) to X(3006)), Peter Moses
noted (Hyacinthos, 12/9/2004) a method for mapping a pair of antipodal
points on one circle to an antipodal pair on another circle. The method
depends on centers of similitude:

Suppose O1 and O2 are circles, that P is on
O1 and that P' is the antipode of P on O1. Let U
be the internal center of similitude (insimilicenter) of O1
and O2, and V the exsimilicenter. Define Q =
PU ∩P' V and Q' = PV ∩P' U. Then on
O2, point Q' is the antipode of Q. Moreover, the lines PP'
and QQ' are parallel.

X(3035) = COMPLEMENT OF X(11)

Let P = X(100) and G=X(2); let GA be the centroid of the
triangle BCP. Define GB and GC cyclically. Then
G, GA, GB, GC are the vertices of a
quadrilateral that is homothetic to the cyclic quadrilateral having
vertices A, B, C, P. The center of homothety is X(3035). Moreover,
X(3035) is the centroid of both quadrilaterals and is the Feuerbach
point of the medial triangle. (Randy Hutson, 9/23/2011)

X(3053) = INTERSECTION X(3)X(6)∩X(4)X(230)

Let Γ1 be the circumcircle, Γ2 the 2nd Lemoine circle, and Γ3 the circle {{X(371),X(372),PU(1),PU(39)}} (which has center X(32)). The three circles intersect in two points, at which let L and L' be the lines tangent to Γ1. Then X(3035) = L∩L'. (Randy Hutson, January 29, 2015)

X(3066) = 2nd LEMOINE HOMOTHETIC
CENTER

Suppose X is a point, with isogonal conjugate X- 1. It is
well known that the pedal triangle of X is homothetic to the antipedal
triangle of X- 1. The Lemoine homothetic center, X(1285), is
the homothetic center when X = X(6), and X(3066) is the homothetic
center when X = X(2). See also X(1285). (Peter Moses, Dec. 7, 2005)

In general if X = x : y : z (trilinears), then the homothetic center
is given by

X(3067) = HOFSTADTER ELLIPSE
INTERSECTION

The Hofstadter ellipse E(0) is described at X(359). The point other
than A, B, and C in which E(0) meets the circumcircle is X(3067). As
with X(359) and X(360), this is a transcendental center, with "exposed
angles" A, B, C in its coordinates.

X(3078) = DANNEELS POINT OF X(5)

Let A'B'C' be the cevian triangle of a point X = x : y : z (trilinears), let L(A) be the line through A parallel to B'C', and define L(B) and L(C) cyclically. The lines L(A), L(B), L(C) form a triangle homothetic to A'B'C', with homothetic center

ax2(by + cz) : by2(cz + ax) : cz2(ax + by).

We denote this point by D(X) and call it the Danneels point of X. (Eric Danneels, Hyacinthos, Jan. 28, 2005). In addition to the discussion below, see the preamble to X(8012) and the Danneels points beginning at X(8012).

If X lies on the line at infinity, then D(X) = X(2) of cevian triangle of X. (Randy Hutson, Jul 23, 2015)

The formula is simpler in barycentrics: if U = u : v : w, then

D(U) = u2(v + w) : v2(w + u) : w2(u + v).

If X is line the Euler line, then D(X) is on
the Euler line. A proof follows. Let a1, b1,
c1 be cos A, cos B, cos C, respectively, so that the Euler
line is given parametrically as x(t) : y(t) : z (t) by

x = b1c1 +
ta1
y = c1a1 +
tb1
z = a1b1 +
tc1,

where t is an arbitrary function homogenous of degree 0 in a, b, c.
Then D(X) is the point U = u : v : w given by

It is easy to check that the point U satisfies this equation. (This
sort of algebraic proof is more inclusive than a geometric proof,
because here, a,b,c are indeterminates or variables. They can, for
example, take values that are not sidelengths of a triangle. Here, cos
A is defined as (b2 + c2 - a2)/(2bc),
so that no dependence on a geometric angle is necessary.)

The appearance of (i,j) in the following list means that D(X(i)) = X(j):
(1,42) (2,2) (3,418)
(4,25) 6,3051 (7,57)
(8,200) (69,394) (75,321)
(100,55) (110,84) (264,324)
(366,367) (651,222) (653,196)
(1113,25) (1114,25) (1370,455) For an extension of this list, see the preamble to X(8012).

Let X be a point on the 2nd Brocard circle. The locus of X(4) of triangle XPU(1) as X varies is a circle with center X(3095). This circle is the reflection of the 2nd Brocard circle in X(39). See also X(9821). (Randy Hutson, July 20, 2016)

The 1st Brocard point (trilinears c/b : a/c : b/a) and 2nd Brocard
point (b/c : c/a : a/b) are opposing vertices of a square here called
the Brocard square. The other two vertices are X(3102) and X(3103).
Both are on the Brocard axis X(3)X(6), and X(3102) is the closer to
X(3).

Then the points A, B, C, gX, tX, Tg, Gt are on a conic. As a
circumconic, it is the image under the isogonal conjugate mapping of
line X(gtX). It is also the image under the isotomic conjugate mapping
of line X(tgX).

Suppose A1B1C1 is the cevian triangle
of a point X.
Let LAB be the line through B parallel to
A1B1, and let LAC be the line through
C parallel to A1C1.
Let A2 = LAB∩LAC, and define
B2 and C2 cyclically. Let

X(3123) = X(i)-Ceva conjugate of X(i) and X(j) for these (i,j):
(76,661), (1015,244)
X(3123) = crosspoint of X(75) and X(513)
X(3123) = crosssum of X(i) and X(j) for these (i,j): (31,100),
(87,932), (101,2209)
X(3123) = crossdifference of every pair of points on the line
X(101,932)

As a point on the Euler line, X(3146) has Shinagawa coefficients (1, -4).

The point X(3146) exemplifies a theorem based on a discussion in Tran Quang Hung and Peter Moses, Hyacinthos 25668.
The theorem (Peter Moses, April 10, 2017) is as follows.

Suppose that P is a point on a circumhyperbola H. Let A'B'C' be the cevian triangle of P and let O' be the circumcircle of A'B'C'. Let A'' be the point other than A' that lies on both O' and line BC, and define B'' and C'' cyclically. The orthocenter of A"B"C" lies on the isogonal conjugate of H.

X(3154) = midpoint of X(125) and X(3258)
X(3154) = inverse-in nine-point circle of X(3134)
X(3154) = X(477)-Ceva conjugate of X(523)
X(3154) = crossdifference of every pair of point on the line
X(647)X(2420)

X(3167) =X(i)-Ceva conjugate of X(j) for these (i,j): (6,3),
(193,3053)
X(3167) = crosspoint of X(6) and X(3053)
X(3167) = crosssum of X(2) and X(2996)
X(3167) = perspector of unary cofactor triangles of outer and inner Vecten triangles
X(3167) = centroid of anticevian triangle of X(3), which is also the antipedal triangle of X(64) and the tangential triangle of the MacBeath circumconic

X(3187) = anticomplement of X(306)
X(3187) = X(27)-Ceva conjugate of X(2)
X(3187) = crosspoint of X(648) and X(1016)
X(3187) = crosssum of X(647) and X(1015)
X(3187) = crossdifference of every pair of points on the line
X(649)X(838)

X(3216) = X(58)-CEVA CONJUGATE OF X(1)

Let A'B'C' be the medial triangle and A''B''C'' then cevian triangle of X(1). Let
U be the circumcircle of A, A', A'', and define V and W cyclically. Then X(3216) is the radical center of U, V, W. See Antreas Hatzipolakis and Peter Moses, Hyacinthos 267292.

X(3221) = SS(a → bc) OF
X(647)

Trilinears for X(3221) are obtained by applying the symbolic
substitution (a,b,c) → (bc,ca,ab) to trilinear coordinates of
X(647). As a symbolic substitution, this mapping takes lines onto
lines. In particular, points on the Euler line, X(2)X(3), with
coefficients given by X(647), are mapped to the line X(6)X(194), with
coefficients given by X(3221). Symbolic substitutions are introduced in
the following article:

X(3225) = SS(a → bc) OF
X(98)

Trilinears for X(3225) are obtained by applying the symbolic
substitution (a,b,c) → (bc,ca,ab) to trilinear coordinates of X(98).
As a symbolic substitution, this mapping takes circumconics to
circumconics. In particular, it maps circumcircle onto the Steiner
circumellipse, which passes through X(3225).

X(3226) = SS(a → bc) OF
X(105)

Trilinears for X(3226) are obtained by applying the symbolic
substitution (a,b,c) → (bc,ca,ab) to trilinear coordinates of
X(105). As a symbolic substitution, this mapping takes circumconics to
circumconics. In particular, it maps circumcircle onto the Steiner
circumellipse, which passes through X(3226).

X(3227) = SS(a → bc) OF
X(106)

Trilinears for X(3227) are obtained by applying the symbolic
substitution (a,b,c) → (bc,ca,ab) to trilinear coordinates of
X(106). As a symbolic substitution, this mapping takes circumconics to
circumconics. In particular, it maps circumcircle onto the Steiner
circumellipse, which passes through X(3227).

X(3228) = SS(a → bc) OF
X(111)

Trilinears for X(3228) are obtained by applying the symbolic
substitution (a,b,c) → (bc,ca,ab) to trilinear coordinates of
X(111). As a symbolic substitution, this mapping takes circumconics to
circumconics. In particular, it maps circumcircle onto the Steiner
circumellipse, which passes through X(3228).

X(3229) = ISOGONAL CONJUGATE OF X(3225)

X(3229) = isogonal conjugate of X(3225)
X(3229) = X(i)-Ceva conjugate of X(j) for these (i,j): (385,511), (699,6)
X(3229) = crosspoint of X(i) and X(j) for these (i,j): (2,694), (6,699)
X(3229) = crosssum of X(i) and X(j) for these (i,j): (2,698), (6,385), (192,2238), (523,2086)
X(3229) = crossdifference of every pair of points on the line X(2)X(669)
X(3229) = complement of X(3978)
X(3229) = crossdifference of PU(148)
X(3229) = intersection of line X(2)X(39)[X(194)] and line through X(2)-Ceva conjugate of X(194) and X(194)-Ceva conjugate of X(2)

Propetries of the inscribed Yff parabola: center = X(514);
perspector = X(190), focus = X(101); directrix = X(4)X(9), the
trilinear pole of X(1897). The parabola passes through these points:
X(514), X(649), X(3234), X(3239). The axis of the parabola is
X(101)X(514), and the Simson line of the focus is X(118)X(516).
Contributed by Peter Moses, April 6, 2007.

X(3248) = X(i)-cross conjugate of X(j) for these (i,j): (3121,1015),
(3249,649)
X(3248) = crosspoint of X(i) and X(j) for these (i,j): (1,649),
(31,667), (873,1019), (1015,1357)
X(3248) = crosssum of X(i) and X(j) for these (i,j): (1,190), (75,668),
(341,646), (872,1018)
X(3248) = crossdifference of every pair of points on the line
X(190)X(646)

X(3249) = SS(a → bc) OF
X(764)

X(3249) = X(i)-Ceva conjugate of X(j) for these (i,j): (649,3248),
(1919,1977)
X(3249) = crosspoint of X(i) and X(j) for these (i,j): (649,3248),
(1919,1977)
X(3249) = crossdifference of every pair of points on the line
X(350)X(899)

X(3250) = INTERSECTION OF X(37)X(513) AND X(187)X(237)

Let L be the line P(6)U(6) = X(37)X(513). Let M be the trilinear polar of the cevapoint of PU(6), this point being X(256). Let V = P(6)-Ceva conjugate of U(6) and W = U(6)-Ceva conjugate of P(6). The lines L, M, VW concur in X(3250). (Randy Hutson, December 26, 2015)

X(3250) = reflection of X(649) in X(665)
X(3250) = X(i)-Ceva conjugate of X(j) for these (i,j): (825,6), (2186,2170)
X(3250) = crosspoint of X(6) and X(825)
X(3250) = crosssum of X(2) and X(824)
X(3250) = crossdifference of every pair of points on the line X(2)X(31)
X(3250) = isogonal conjugate of X(4586)

X(3251) = INTERSECTION OF X(1)X(513) AND
X(42)X(663)

X(3251) = reflection of X(1635) in X(1960)
X(3251) = X(i)-Ceva conjugate of X(j) for these (i,j): (1,2087),
(100,44)
X(3251) = crosspoint of X(i) and X(j) for these (i,j): (1,1023),
(44,100)
X(3251) = crosssum of X(i) and X(j) for these (i,j): (1,1022),
(88,513), (100,3257)
X(3251) = crossdifference of every pair of points on the line
X(44)X(88)

X(3257) = ISOGONAL CONJUGATE OF X(1635)

Let P and Q be the intersections of line BC and circle {X(3),2R}. Let A' be the circumcenter of triangle PQX(3), and define B' and C' cyclically. The lines AA', BB', CC' concur in X(3257). (Randy Hutson, December 26, 2015)

X(3258) = CROSSSUM OF X(74) AND
X(110)

Let P = X(477) and H=X(4); let HA be the orthocenter of
the triangle BCP. Define HB and HC cyclically.
Then H, HA, HB, HC are the vertices of
a quadrilateral that is homothetic to the cyclic quadrilateral having
vertices A, B, C, P. The center of homothety is X(3258). Moreover,
X(3258) is the anticenter of the quadrilateral ABCP. (Randy Hutson,
9/23/2011)

Let La, Lb, Lc be the lines through A, B, C, resp. parallel to the Euler line. Let Ma, Mb, Mc be the reflections of BC, CA, AB in La, Lb, Lc, resp. Let A' = Mb∩Mc, and define B' and C' cyclically. Triangle A'B'C' is inversely similar to, and 3 times the size of, ABC. Let A"B"C" be the reflection of A'B'C' in the Euler line of A'B'C' (line X(30)X(74)). The triangle A"B"C" is homothetic to ABC, with center of homothety X(3258); see Hyacinthos #16741/16782, Sep 2008. (Randy Hutson, March 25, 2016)

X(3259) = midpoint of X(4) and X(953)
X(3259) = complement of X(901)
X(3259) = complementary conjugate of X(900)
X(3259) = X(i)-Ceva conjugate of X(j) for these (i,j): (2,3310), (4,900), (11,1647)
X(3259) = crosssum of X(100) and X(104)
X(3259) = crossdifference of every pair of points on the line X(2423)X(2427)
X(3259) = inverse-in-polar-circle of X(1309)
X(3259) = perspector of circumconic centered at X(3310)
X(3259) = center of circumconic that is locus of trilinear poles of lines passing through X(3310); this conic is a rectangular circumhyperbola that is isogonal conjugate of line X(3)X(8)

The points X(3272) to X(3283) are associated with six equilateral
triangles related to the (1st) Morley triangle; trilinears for vertices
of these triangles are given just below. Received from Milorad R.
Stevanovic, Nov. 24, 2007 and Dec. 25, 2007.

Circumtangential triangle, inscribed in the circumcircle of
ABC and homothetic to the 1st Morley triangle,
T1T2T3. For trilinears, of the
vertices, see TCCT, page 166, or MathWorld. Trilinears found by M.
Stevanovic:

Circumnormal triangle inscribed in the circumcircle of ABC
and homothetic to the 1st Morley triangle,
N1N2N3. For trilinears, of the
vertices, see TCCT, page 166, or MathWorld. Trilinears found by M.
Stevanovic:

X(3288) = reflection of X(i) in X(j) for these (i,j): (2451,3049),
(3049,3050)
X(3288) = crosspoint of X(99) and X(3114)
X(3288) = crosssum of X(512) and X(3117)
X(3288) = crossdifference of every pair of points on the line
X(2)X(51)

X(3289) = reflection of X(3331) in X(1625)
X(3289) = X(i)-Ceva conjugate of X(j) for these (i,j): (287,3),
(2421,684)
X(3289) = crosspoint of X(3) and X(287)
X(3289) = crosssum of X(4) and X(232)
X(3289) = crossdifference of every pair of points on the line
X(4)X(512)

X(3290) = INTERSECTION OF LINES X(2)X(37) AND X(230)X(231)

Trilinears b3 + c3 - 2abc + (a2 - bc)(b + c) : :

Let L denote the line through X(3) perpendicular to the line X(1)X(6). Coefficients for a trilinear equation for L are the
trilinears for X(3290).

X(3291) = midpoint of X(3124) and X(3231)
X(3291) = complement of X(3266)
X(3291) = X(i)-Ceva conjugate of X(j) for these (i,j): (2,126),
(892,512)
X(3291) = crosssum of X(i) and X(j) for these (i,j): (6,524),
(525,1648)
X(3291) = crossdifference of every pair of points on the line
X(3)X(669)

X(3292) = INTERSECTION OF LINES X(3)X(49) AND
X(23)X(110)

Let L denote the line through X(4) perpendicular to the line
X(2)X(6). Coefficients for a trilinear equation for L are the
trilinears for X(3292).

Let O* be the circle with segment X(15)X(16) as diameter (and center
X(187). Let P be the perspector of O*. Then X(3292) is the trilinear
pole of the polar of P with respect to O*. See X(5642) for a similar
property involving the segment X(13)X(14). (Randy Hutson, July 18,
2014)

X(3292) = midpoint of X(110) and X(323)
X(3292) = reflection of X(1495) in X(1)
X(3292) = X(i)-Ceva conjugate of X(j) for these (i,j): (524,187),
(895,3)
X(3292) = crosspoint of X(3) and X(895)
X(3292) = crosssum of X(4) and X(468)
X(3292) = crossdifference of every pair of points on the line
X(4)X(1499)

Suppose P and U are points on the line at infinity, given in
barycentric coordinates by P = p : q : r and U = u : v : w. We call P
and U perpendicular directions if for every point X not in the
line at infinity, the lines XP and XU are perpendicular. (It is well
known that if P and U are an antipodal pair on the circumcircle, then
their isogonal conjugates are perpendicular directions.)

Theorem: If P and U are perpendicular directions, then their
barycentric product lies on the orthic axis.

An outline of a proof follows. For given P on the line at infinity,
the direction perpendicular to P is given by

For P to lie on the line at infinity means that r = - p - q.
Substitution for r, a computer quickly shows that

x cot A + y cot B + z cot C = 0,

as desired.

The orthic axis is perpendicular to the Euler line. For this and
other properties, visit MathWorld. The appearance of (I, J, K) in the
following list means that X(K) is the barycentric product of
perpendicular directions X(i) and X(j).

X(3311) is the perspector of each of the following pairs of triangles:
Lucas central triangle and the symmedial triangle (the cevian triangle of X(6))
Lucas tangents triangle and the Lucas(-1:1) central triangle
Lucas(2:3) central triangle and the circumsymmedial triangle.
Moreover, X(3311), is the radical center of the Lucas(4:1) circles. See X(371) and X(3312). (Randy Hutson,
9/23/2011)

X(3311) is the perspector of each of the pair of the following four triangles:
symmedial triangle. Lucas central triangle, Lucas(-1) secondary central triangle, 1st Lucas(-1) secondary tangents triangle. (Randy Hutson, September 14,. 2016)

X(3312) is the perspector of each of the following pairs of triangles:
Lucas(-1:1) central triangle and the symmedial triangle
Lucas(-1:1) tangents triangle and the Lucas central triangle
Lucas(-2:3) central triangle and the circumsymmedial triangle.
Moreover, X(3311), is the radical center of the Lucas(-4:1) circles. See X(371) and X(3311). (Randy Hutson, 9/23/2011)

X(3312) is the perspector of each of the pair of the following four triangles:
symmedial triangle, Lucas(-1) central triangle, Lucas secondary central triangle, 1st Lucas secondary tangents triangle (Randy Hutson, September 14, 2016)

which lies on the incircle. Geometrically, if U is not X(1) then T(U)
is the perspector of ABC and the reflection of the intouch triangle in
the line UX(1). If X is a point other than U and X(6), then T(X) = T(U)
if and only if X lies on the line UX(6). The transform T is clearly
closely related to the Brisse transform, described just before X(1354).

X(3327) lies on the incircle and these lines:
12,128 55,930 56,1141 496,1263

X(3227) = Steiner-circumellipse-antipode of X(668)
X(3227) = projection from Steiner inellipse to Steiner circumellipse of X(1015)
X(3227) = the point of intersection, other than A, B, and C, of the Steiner circumellipse and hyperbola {{A,B,C,X(1),X(2)}}
X(3227) = antipode of X(2) in hyperbola {{A,B,C,X(1),X(2)}}
X(3227) = trilinear pole of line X(2)X(513)

X(3328) lies on the incircle and these lines:
55,1308 56,2717 516,1317
517,1362

X(3328) = reflection of X(i) in X(j) for these (i,j): (1155,1323), (3322,1)
X(3328) = X(7)-Ceva conjugate of X(1638)
X(3328) = crosspoint of X(i) in X(j) for these (i,j): (7,1638), (514,1323)

X(3228) = Steiner-circumellipse-antipode of X(670)
X(3228) = projection from Steiner inellipse to Steiner circumellipse of X(1084)
X(3228) = the point of intersection, other than A, B, and C, of the Steiner circumellipse and hyperbola {{A,B,C,X(2),X(6)}}
X(3228) = antipode of X(2) in hyperbola {{A,B,C,X(2),X(6)}}
X(3228) = trilinear pole of line X(2)X(512)

X(3333) = POHOATA
POINT

Let I be the incenter, X(1), and let KA be the symmedian
point of triangle IBC; define KB, KC cyclically.
Let X be the midpoint of segment AI,and define Y, Z cyclically. Then
the triangles KAKBKC and XYZ are
perspective, and their perspector is X(3333). Contributed by Cosmin
Pohoata, April 4, 2008.

Let R be the circumradius, r the inradius, and rA the
radius of the A-excircle. The trilinears given above are equivalent to
2R - rA : 2R - rB : 2R - rC. The 1st
trilinear representation

(a - b + c)(a + b - c)(a + b + c) + 4abc

shows that X(3333) lies on the line IO of the incenter and the
circumcenter, which is also the line X(1)X(57); viz., X(57) has 1st
trilinear (a - b + c)(a + b - c) and X(1) has 1st trilinear 1, so that
the above representation may be viewed as a linear combination using as
coefficients the symmetric functions a + b + c and 4abc. The
representation generalizes to

(a - b + c)(a + b - c)(a + b + c) + kabc,

where k is a symmetric function of degree 0 in a,b,c. Every point on
the line IO necessary has such a form. (Indeed, every point on every
line is given by such linear combinations of any two points on the
line.) For the line IO, Peter Moses contributed (April 24, 2008) a list
of points with corresponding functions k(a,b,c). Aside from inradius r
and circumradius R, other symbols used are identified elsewhere in ETC
as indicated, or else are formulated here: s = (a+b+c)/2; J as at
X(1113); e as at X(1340); SA, SB SC as
at X(1117); and sA = (-a+b+c)/2, sB = (a-b+c),
sC = (a+b-c)/2.

X(3334) = PERSPECTOR OF TRIANGLES
T1T2T3 AND
J1J2J3

The equilateral triangles T1T2T3
and J1J2J3 are defined just before
X(3272). The points X(3334) and X(3335), contributed April 16, 2008 by
Peter Moses, complete a list (just before X(3272) of perspectors
associated with equilateral triangles.

X(3361) = INTERSECTION OF LINES X(1)X(3) AND
X(7)X(1125)

Let OA be the circle tangent to side BC at its midpoint
and to the circumcircle on the side ob BC opposite A. Define
OB and OC cyclically. Let A' be the
insimilicanter of OB and OC, and define B' and C'
cyclically. The lines AA', BB', CC' concur in X(3361). See the
reference at X(1001).

X(3363) = X(6) OF PEDAL TRIANGLE OF X(2)

As a point on the Euler line, X(3363) has Shinagawa coefficients
(2(E + F)2 + 3S2,9S2).

X(3363) is the symmedian point of the pedal triangle of the
centroid. A second construction uses the midpoint X(597) of the
symmedian point and centroid and the midpoint X(115) of the two Fermat
points, X(13) and X(14); specifically, X(3363) is the point in which
the line X(115)X(597) meets the Euler line. Contributed by Po-chieh
Chen and Shao-cheng Liu, May 29, 2008.

X(3364) = COS(A - 5π/12)
POINT

X(3364) and related points occupy Bernard Gibert's Table 38 concerning these three loci: Brocard
axis, Kiepert hyperbola, and the cubic K457. Trilinear equations for
these curves, of degrees 1,2,3, respectively, in the variables a,b,c,
are given as follows:

For a geometric interpretation, let T be the
vertex triangle of the circumcevian triangles,
AUBUCU and
AXBXCX, of U and X; viz., the
sidelines of T are AUAX,
BUBX, CUCX. Then T is
perspective to ABC, and the perspector is the U-vertex conjugate of
X.

The definition of vertex conjugate allows X = U. To extend the
geometric interpretation to the case that X = U, as X approaches U, the
vertex triangle approaches a limiting triangle which we call the
tangential triangle of U, a triangle perspective to ABC with perspector
U-vertex conjugate of U.

The appearance of a row I, J, K in the following tables signifies
that the X(i)-vertex conjugate of X(j) is X(K).

I

J

K

1

1

56

1

2

3415

1

3

84

1

4

3417

1

6

2163

1

7

3418

1

9

3420

1

19

3422

1

56

1

1

57

3423

1

58

58

2

2

25

2

3

3424

2

4

3425

2

6

1383

2

32

3407

2

251

251

2

523

23

3

3

64

3

4

4

3

6

3426

3

7

3427

3

20

3346

3

40

3345

3

56

945

3

64

3

3

84

1

3

1490

3447

3

1498

3348

3

2131

3183

3

3182

3354

4

4

3

5

5

3432

6

6

6

7

7

3433

8

8

3434

9

9

1436

10

10

3437

Among properties of vertex conjugation are these:

1. X(3)-vertex conjugation maps the Darboux cubic to the Darboux
cubic. The appearance of (i,j) in the following list means that X(i) is
on the Darboux cubic and that X(j) = X(3)-vertex conjugate of X(i):

2. The fixed point of U-vertex conjugation is
the 1st Saragossa point of U. (Saragossa points are defined just before
X(1166).) The appearance of (i,j) in the following list means that the
1st Saragossa point of X(i) is X(j):

X(3447) = X(523)-VERTEX CONJUGATE OF X(523)

Let E be the Euler line and TATBTC the tangential triangle of ABC. Let DA = E∩TBTC, and define DB and DC cyclically. Let A' = DBTB∩DCTC, and define B' and C' cyclically. Then A'B'C' is perspective to ABC, and the perspector is X(3447). For a sketch, click X(3447)andX(7669). (Angel Montesdeoca, April 22, 2016)

Let A'B'C' be the orthocentroidal triangle and A"B"C" the anti-orthocentroidal triangle.
Let A* be the reflection of A" in B'C', and define B* and C* cyclically. The lines AA*, BB*, CC* concur in X(74), and X(3448) = centroid of A*B*C*. (Randy Hutson, December 10, 2016)

X(3465) = reflection of X(484) in X(2222)
X(3465) = isogonal conjugate of X(3466)
X(3465) = X(i)-Ceva conjugate of X(j) for these (i,j): (30,484),
(1807,1)
X(3465) = crosssum of X(654) and X(2638)
X(3465) = crossdifference of every pair of points on the line
X(652)X(2260)

X(3487) = INTERSECTION OF LINES X(1)X(4) AND X(3)X(7)

Trilinears 1 + cos A + cos B + cos C + cos B cos C : :

Let A' be the midpoint of X(1) and the A-intouch point. Define B' and C' cyclically. The triangle A'B'C' is homothetic to the 2nd extouch triangle, and the center of homothety is X(3487). (Randy Hutson, September 14, 2016)

X(3509) = isogonal cojugate of X(3512)
X(3509) = X(i)-Ceva conjugate of X(j) for these (i,j): (335,1),
(385,3511)
X(3509) = crosssum of X(i) and X(j) for these (i,j): (659,2170),
(2238,2292)
X(3509) = crossdifference of every pair of points on the line
X(663)X(1193)

for a discussion of X(3513) as a point of intersection of three similar ellipses, each having two of the points A, B, C as foci. (Contributed by David Eppstein, 7/10/08)

X(3513) is the point P such that the incenter of the circumcevian
triangle of P is also the incenter of triangle ABC. (Randy Hutson,
9/23/2011)

X(3513) and X(3514) are the limiting points (point-circles) of the
coaxal system that includes the circumcircle and incircle. Their
midpoint, X(3660), is the radical trace of the incircle and
circumcircle. (Peter Moses, November 15, 2011)

X(3519) = ISOGONAL CONJUGATE OF
X(3518)

Let A'B'C' be the reflection triangle. Let Oa be the circle centered at A' and passing through A, and define Ob and Oc cyclically. The radical center of Oa, Ob, Oc is X(3519).
Let Na be the reflection of X(5) in the perpendicular bisector of BC, and define Nb and Nc cyclically. The lines ANa, BNb, CNc concur in X(3519). Let A''B''C'' be the Trinh triangle. Let A* be the orthopole of line B'C', and define B* and C* cyclically. The lines AA*, BB*, CC* concur in X(3519). (Randy Hutson, October 13, 2015)

As a point on the Euler line, X(3522) has Shinagawa coefficients (3, -4).

X(3522) is one of many points on the Euler line that have a certain
convenient representation, received from Peter Moses, February 8, 2010.
Specifically, the appearance of (t,k) in the following list means
that

X(3548) = EULER LINE INTERCEPT OF LINE X(68)X(125)

Trilinears sec B sec C - 4 sin B sin C : :

As a point on the Euler line, X(3548) has Shinagawa coefficients (E - 4F, -E).

Let La be the polar of X(4) wrt the circle centered at A and passing through X(3), and define Lb and Lc cyclically. (Note that X(4) is the perspector of any circle centered at a vertex of ABC.) Let A" = Lb∩Lc, and define B" and C" cyclically. Triangle A"B"C" is homothetic to the medial triangle, and the center of homothety is X(3548). (Randy Hutson, December 11, 2015)

X(3557) was submitted and called the Pappus point by Roland Bacher
(March 11, 2009); coordinates and the related point X(3558) were found
by Peter Moses (March 12, 2009).

X(3557) is the crosspoint (and crosssum) of the real foci of the Steiner inellipse. (Bernard Gibert, January 4, 2015)

X(3557) is the Brocard axis intercept, other than X(1340), of the circle {{X(1340),PU(1)}}. Also, X(3557) is the insimilicenter of the 2nd Lemoine circle and the circle {{X(371),X(372),PU(1),PU(39)}}. (Randy Hutson, January 5, 2015)

X(3558) = 2nd PAPPUS POINT

X(3558) is the crosspoint (and crosssum) of the imaginary foci of the Steiner inellipse. (Bernard Gibert, January 4, 2015)

X(3558) is the Brocard axis intercept, other than X(1341), of the circle {{X(1341),PU(1)}}. Also, X(3558) is the exsimilicenter of the 2nd Lemoine circle and the circle {{X(371),X(372),PU(1),PU(39)}}. (Randy Hutson, January 5, 2015)

X(3563) = 1st MOSES CIRCUMCIRCLE
POINT

The antipodal pair X(3563) to X(3565) were described by Peter Moses, Dec. 13, 2004.

Let A' be the reflection in line BC of the A-vertex of the antipedal triangle of X(6), and define B' and C' cyclically. Let OA be the circumcenter of AB'C', and define OB and OC cyclically. Let U be the circumcenter of A'BC, and define V and W cyclically. Let O' be the circumcenter of OAOBOC. Let O'' be the circumcenter of UVW. Then X(3563) = Λ(O',O''). Also, O'O''∩(infinity line) = X(3564). (Randy Hutson, June 19, 2015)

X(3563) = reflection of X(3565) in X(3)
X(3563) = isogonal conjugate of X(3564)
X(3563) = cevapoint of X(25) and X(232)
X(3563) = Λ(X(5), X(6))
X(3563) = the point of intersection, other than A, B, and C, of the circumcircle and hyperbola {A,B,C,X(2),X(24)}
X(3563) = the point of intersection, other than A, B, and C, of the circumcircle and hyperbola {A,B,C,X(4),X(25)}
X(3563) = inverse-in-polar-circle of X(114)
X(3563) = inverse-in-{circumcircle, nine-point circle}-inverter of X(136)
X(3563) = trilinear pole of line X(6)X(924)

X(3564) = 1st MOSES INFINITY POINT

X(3564) and X(3566) are on the line at infinity. They may be
regarded as directions in the plane of ABC, and as such are
perpendicular.
Continuing from X(3563), the infinite point of the line O'O'' is X(3564). (Randy Hutson, June 19, 2015)

Let P(2) and U(2) be the 1st and 2nd Beltrami points (as indexed at
Bicentric Pairs, accessible using the Tables button at the top of ETC),
and let P(40) and U(40) be the isogonal conjugates of P(2) and U(2),
respectively. Then X(3568) is the point of intersection of the lines
P(2)P(40) and U(2)U(40). The name "Beltrami-Euler Point" signifies the
fact that X(3568) lies on the Euler line. See also X(2966) and X(3569).
Contributed by Chris van Tienhoven, June 7, 2009.

X(3568) lies on this line: 2,3

X(3569) = BELTRAMI-PARRY POINT

Let P(2) and U(2) be the 1st and 2nd Beltrami points (as indexed at
Bicentric Pairs, accessible using the Tables button at the top of ETC),
and let P(40) and U(40) be the isogonal conjugates of P(2) and U(2),
respectively. Then X(3569) is the point of intersection of the lines
P(2)U(2) and P(40)U(40). The name "Beltrami-Parry Point" signifies the
fact that X(3569) lies on the line X(74)X(111); where X(111) is the
Parry point.and X(74) is the isogonal conjugate of the Euler infinity
point. See also X(3568). Contributed by Chris van Tienhoven, June 7,
2009..

X(3570) = INTERSECTION OF LINES P(6)U(8) AND P(8)U(6)

The points given by trilinears P(6) = b : c : a and U(6) = c : a : b
are indexed at Bicentric Pairs (accessible using the Tables button at
the top of ETC), and likewise for their isogonal conjugates P(8) = 1/b
: 1/c : 1/a and U(8) = 1/c : 1/a : 1/b. The bicentric pair of lines
P(6)U(8) and P(8)U(6) concur in X(3570). Contributed by Peter Moses,
July 7, 2009..

X(3571) = INTERSECTION OF LINES P(6)P(8) AND U(6)U(8)

The points given by trilinears P(6) = b : c : a and U(6) = c : a : b are indexed at Bicentric Pairs (accessible using the Tables button at the top of ETC), and likewise for their isogonal conjugates P(8) = 1/b : 1/c : 1/a and U(8) = 1/c : 1/a : 1/b. The bicentric pair of lines P(6)P(8) and U(6)U(8) concur in X(3571). Contributed by Peter Moses, July 7, 2009.

Continuing, the line P(6)P(8) is the tangent to the 1st bicentric of the Kiepert hyperbola at P(8), and line U(6)U(8) is the tangent to the 2nd bicentric of the Kiepert hyperbola at U(8). (Randy Hutson, March 25, 2016)

X(3571) lies on these lines: 1,512 9,43
1621,1964

X(3571) = crossdifference of PU(90)
X(3571) = crossdifference of every pair of points on line X(2238)X(4367)

X(3572) = INTERSECTION OF LINES P(6)U(6) AND
P(8)U(8)

The points given by trilinears P(6) = b : c : a and U(6) = c : a : b
are indexed at Bicentric Pairs (accessible using the Tables button at
the top of ETC), and likewise for their isogonal conjugates P(8) = 1/b
: 1/c : 1/a and U(8) = 1/c : 1/a : 1/b. The pair of central lines
P(6)U(6) and P(8)U(8) concur in X(3572). Contributed by Peter Moses,
July 7, 2009.

X(3575) = EULER LINE INTERCEPT OF THE LINE
X(64)X(66)

As a point on the Euler line, X(3575) has Shinagawa coefficients (F, -E - 3F).

Let KH denote the hyperbola discussed at X(1112). As noted in the
paper cited at X(1112), KH is the X(4)-Ceva conjugate of the Euler
line. Inversely, the Euler line is the X(4)-Ceva conjugate of KH. Since
X(3574) lies on KH, its X(4)-Ceva conjugate, which is X(3575), lies on
the Euler line. (Peter Moses, July 7, 2009)

X(3580) = INTERSECTION OF LINES X(2)X(6) AND X(30)X(74)

Let La, Lb, Lc be the lines through A, B, C, resp. parallel to the orthic axis. Let Ma, Mb, Mc be the reflections of BC, CA, AB in La, Lb, Lc, resp. Let A' = Mb∩Mc, and define B' and C' cyclically. Triangle A'B'C' is inversely similar to, and 3 times the size of, ABC. Let A"B"C" be the reflection of A'B'C' in the orthic axis. The triangle A"B"C" is homothetic to ABC, and its centroid is X(3580); see Hyacinthos #16741/16782, Sep 2008. (Randy Hutson, March 25, 2016)

X(3583) = reflection of X(i) in X(j) for these (i,j): (36,11), (484,1737)
X(3583) = crosspoint of X(79) and X(80)
X(3583) = crosssum of X(35) and X(36)
X(3583) = homothetic center of 2nd isogonal triangle of X(1) and the reflection of the Johnson triangle in X(4); see X(36)
X(3583) = homothetic center of Mandart-incircle triangle and (cross-triangle of ABC and 2nd isogonal triangle of X(1))

In 2008, Alexei Myakishev gave the following construction of a
conic. Let CA and CB be points on the line AB
satisfying |BCA| = |CB| and |ACB| = |CA| and
arranged in this order: CA, B, A, and CB. Define
points AB, CB, BC, AC
cyclically. The six points CA, BA, AB,
CB, BC, AC lie on a conic. Myakishev's
proof is by Carnot's theorem, since

X(3589) = ISOGONAL CONJUGATE OF X(3108)

In 2003, Peter Moses gave a general construction as follows: Suppose
that P = u : v : w (barycentrics). The centroids of the triangles BCP,
CAP, ABP form a triangle homothetic to ABC, with ratio -1/3 and
center

P' = 2u + v + w : u + 2v + w : u + v +2w.

In 2010, Seiichi Kirikami gave another construction for P': let

D = AP∩BC, E =
BP∩CA, F = CP∩AB.

Then the Newton lines of the quadrilaterals PEAF, PFBD, PDCE concur
in P'. (The Newton line of a quadrilateral is the line of the midpoints
of the two diagonals of the quadrilateral.)

X(3589) is the Moses-Kirikami image of the symmedian point. In the
following list (from P. Moses, Aug. 23, 2010), the appearance of I,
J means that X(j) is the Moses-Kirikami image of X(i).

Early in 2012, Seiichi Kirikami found a simple relation between the
tetrahedron and the Moses-Kirikami image. Using 3-dimensional cartesian
coordinates, suppose that the vertices of triangle ABC are placed in
the xy-plane:

A = (x1, y1, 0), B = (x2,
y2, 0), C = (x3, y3, 0).

Let P = (x4, y4, z4) be an
arbitrary point not in the plane of ABC. Then the centroid of the
tetrahedron ABCP is the point

Let A' be the midpoint of segment AX(6), and define B' and C' cyclically. Then A'B'C' is homothetic to the medial triangle, and the homothetic center is X(3589). (Randy Hutson, December 26, 2015)

Let E be the bicevian conic of X(2) and X(6); i.e., the ellipse that passes through the vertices of the medial and symmedial triangles. Then X(3589) is the center of E. This ellipse is also the locus of centers of circumconics passing through X(6). Let L be a line through X(2), and let P and P' be the points of intersection of L and the circumcircle. Let V be the locus of the crosssum of P and P'. The locus of V generated by L is E. (Randy Hutson, December 26, 2015)

Let A' be the reflection of X(6) in line BC. Let Oa be the circle with center A' and tangent to BC. Define Ob and Oc cyclically. The radical center of Oa, Ob, Oc is X(3589). (Randy Hutson, December 26, 2015)

On Dec. 22, 2000, Atul Dixit constructed this point in Hyacinthos
message #2183, as follows. Let ABC be a triangle with medians AD, BE,
FC and centroid G. Construct semicircles with diameters BD, DC, BC
outwardly. Let TA be the circle tangent to the three
semicircles BD, DC, BC, and let A' be the center of TA.
Define B' and C' cyclically. Then the lines AA', BB', CC' concur in
X(3590). Barycentric coordinates were found by Paul Yiu (2000), and
further properties and related points, by Peter Moses (2010).

The radius of circle TA is |BC|/6, and the Kiepert angle
(e.g., B-to-C-to-X(3890)) is arctan(2/3).

X(3596) = 1st ODEHNAL
POINT

Let A', B', C' denote the respective excircles of a triangle ABC.
Let A'' be the circle tangent to A', B', C' whose interior includes A',
and likewise for circles B'' and C''. Let Kaa be the point
of tangency of circles A' and A'', and likewise for Let Kbb
and Let Kcc. The lines A-to-Kaa, B-to-Let
Kbb, C-to-Let Kcc concur in X(3596). For more,
see

X(3597) = 2nd ODEHNAL POINT

Let MA, MB, MC be the centers of
the circles A'', B'', C'' used to construct X(3596), respectively. Then
the lines A-to-MA, B-to-MB, C-to-MC
concur in X(3597). For more, see the reference and sketch at
X(3596).

Let OA be the circle passing through B and C, internally
tangent to the incircle, and likewise for OB and
OC. Let PA be the point where OA meets
the incircle, and likewise for PB and PC. Let
QA be the point of intersection of the tangents to the
incircle at PB and PC, and likewise for
PB and PC. Then X(3598) is the perspector of the
triangles PAPBPC and
QAQBQC; this point is also
X(7)-of-QAQBQC. This point and X(3599)
were discovered by Kang-Ying Liu of St. Andrew's Priory School,
Honolulu, Hawaii, during 2010.

Points X(3602)-X(3609) are associated with
the three Morley Cubics indexed as K29, K30, K31 at Bernard Gibert's
Cubics in
the Triangle Plane. The points are related to those in the
section just before X(3272), under the heading "Points Associated with
Equilateral Triangles." See also

X(3609) = 2nd MORLEY-GIBERT
PERSPECTOR

X(3609) is the perspector of the triangle of the cusps of the
Steiner deltoid and the equilateral triangle J defined just before
X(3272). (Bernard Gibert, November 3, 2010)

X(3609) lies on these lines: 5,3272 20,3334
631,3335

X(3610) = 1st AYME-MOSES PERSPECTOR

Trilinears bc(b + c)(a2 - b2 - c2)(a2 + b2 + c2 + 2bc) : :

In a Hyacinthos message dated January 10, 2011, Jean-Louis Ayme
introduced a triangle as follows. Let RA be the radical axis
of the circumcircle and the A-excircle, and define RB and
RC cyclically. Let TA =
RB∩RC, and define TB and
TC cyclically. (TA is also the radical center of
the circumcircle and the B- and C- excircles.) The Ayme triangle
TATBTC is perspective to triangle ABC
and also perspective to many other triangles. Peter Moses found that
its perspector with the cevian triangle of X(346) is X(3610). He also
found that the A-vertex of the Ayme triangle has first barycentric as
follows:

- (b + c)(a2 + b2 + c2 + 2bc) :
b(a2 + b2 - c2) : c(a2 -
b2 + c2),

from which the other two vertices are easily obtained. The Ayme
triangle is perspective to ABC with perspector X(19).

Moses found that the locus of X such that the cevian triangle of X
is perspective to the Ayme triangle is a cubic which passes through the
points X(i) for i = 1, 2, 19, 75, 279, 304, 346, 2184. A barycentric
equation for this Ayme-Moses cubic follows:

(Cyclic sum of ayz[by(a2 + b2 - c2)
- cz(a2 - b2 + c2] ) = 0.

The Ayme triangle is homothetic to the incentral triangle, and the center of homothety is X(612). (Randy Hutson, September 14, 2016)

X(3610) lies on these lines: 10,37 19,346
612,2345

X(3610) = perspector of ABC and cross-triangle of ABC and Ayme triangle

X(3611) = 2nd AYME-MOSES PERSPECTOR

In a Hyacinthos message dated January 7, 2011, Jean-Louis Ayme noted
that the orthic triangle of ABC is perspective to the medial triangle
of the extangents triangle of ABC. Peter Moses found coordinates for
the perspector, X(3611).

X(3613) = ISOTOMIC CONJUGATE OF X(1078)

Let A' be the point of intersection of the tangents to the
nine-point circle at the points where the circle meets line BC, and
define B' and C' cyclically. Then the lines AA', BB', CC' concur in
X(3613). Also, A'B'C' is the side-triangle of the tangential triangle
of the medial triangle and the tangential triangle of the orthic
triangle. (Randy Hutson, August 30, 2011.)

X(3613) is the pole of the Lemoine axis with respect to the
nine-point circle. (Luis González, Hyacinthos #20253, October 5,
2011)

X(3613) is the perspector of the nine-point circle and lies on the
hyperbola that passes through the points A, B, C, X(4), X(5). (Randy
Hutson, December 30, 2012.)

Let A'B'C' be the Feuerbach triangle, L the line through A and X(5),
and A'' = L∩B'C'; define B'' and C'' cyclically. Then the lines
A'A'', B'B'', C'C'' concur in X(3614). X(3614) = {X(5),X(12)}-harmonic
conjugate of X(11). Further, X(3614) is the trilinear pole (with
respect to the Feuerbach triangle) of the perspectrix of ABC and the
Feuerbach triangle. (Randy Hutson, August 30, 2011.)

Suppose that X is a triangle center, that M
is the medial triangle, and that t is a real number. The t-dilation of
M from X, denoted by H(X; M, t), is a triangle center. If X = x : y : z
(trilinears), then

X(3616) = H(X(1); M, 1/2)

Let OA be the circle tangent to side BC at its midpoint
and to the circumcircle on the side of BC opposite A. Define
OB and OC cyclically. Let LA be the
external tangent to circles OB and OC that is the
nearer of the two to OA. Define LB and
LC cyclically. Let A' = LB∩LC, and
define B' and C' cyclically. Then A'B'C' is homothetic to ABC, and the
center of homothety is X(3616). See the reference at X(1001).

In the plane of triangle ABC, let X,Y,Z be the points of contact of
the incircle with sidelines BC,CA,AB, and let X',Y',Z' be the harmonic
conjugates of X,Y,Z with respect to {B,C}, {C,A}, {A,B}, respectively.
It is well known that AX, BY, CZ concur in the Gergonne point, X(7),
and that X',Y',Z' lie on the Gergonne line. The circles with centers
X',Y',Z' passing respectively through X,Y,Z are orthogonal to the
incircle and to the Gergonne line, forming a coaxal system with the
Gergonne line as line of centers. The radical axis is the Soddy line,
which passes through X(1) and X(7) and is the line of centers of the
orthogonal coaxal system, which includes the Soddy circles. Among the
points on this line are the two limiting points (point-circles of the
system): X(3638) and X(3639). The former lies between X(7) and the
inner Soddy point, X(175). (Richard Guy, September 2, 2011)

Peter Moses (November 1, 2010) found that the six points lie on an
ellipse, E, here called the Kirikami-Moses ellipse. The center of E is
the incenter of ABC, and an equation for E is as follows:
e(a,b,c,x,y,z) + e(b,c,a,y,z,x) + e(c,a,b,z,x,y) = 0, where

For any point P, let L(A) be the line through P parallel to BC. Let
U = AB∩L(A) and V = AC∩L(A), and let a' = |UV|. Define b' and
c' cyclically. Let L be the line consisting of points P such that a +
ta' = b + tb' = c + tc'. Then X(3644) is given by t=1. The points
corresponding to t = -1, -1/2, and 0, are X(75), X(37), and X(192),
respectively. (Seiichi Kirikami, November 7, 2010)

X(3644) = reflection of X(i) in X(j) for these (i,j): (75,192),
(1278,37)

Denote the incenter and excenters by I, IA,
IB, IC. Let KA be the centroid of
triangle BCI, and define KB and KC cyclically.
The lines IAKA, IBKB,
ICKC concur in X(3646).

Of the 2 intersections of the Bevan circle and line BC, let Ab be the one closer to B, and define Bc and Ca cyclically. Let Ac be the one closer to C, and define Ba and Cb cyclically. Let Ab' = {B,C}-harmonic conjugate of Ab, and define Bc' and Ca' cyclically. Let Ac' = {B,C}-harmonic conjugate of Ac, and define Ba' and Cb' cyclically. The points Ab', Ac', Bc', Ba', Ca', Cb' lie on an ellipse centered at X(3646). (Randy Hutson, December 10, 2016)

Suppose that X is a point and A'B'C' is a
central triangle. Let LA be the line through A' parallel to
the Euler line of triangle BCX, let LB be the line through
B' parallel to the Euler line of CXA, and let LC be the line
through C' parallel to the Euler line of AXB.

It is well known that if X=X(1), the incenter, then the three
aforementioned Euler lines concur in the Schiffler point, X(21). If
their parallels, the lines LA, LB, LC
concur, the point of concurrence is the Kirikami-Schiffler point of the
triangle A'B'C', denoted by KS(A'B'C'). Seiichi Kirikami (February 1,
2011) found that those lines concur if A'B'C' is the reference triangle
ABC and also concur if A'B'C' is the medial triangle. Peter Moses found
additional cases and properties. A summary follows:

triangle A'B'C'

LA∩LB∩LC

ABC

X(79)

medial

X(3647)

excentral

X(191)

anticomplementary

X(3648)

intouch

X(3649)

extouch

X(3650)

Feuerbach

X(442)

Fuhrmann

X(191)

1st circumperp

X(3651)

2nd circumperp

X(21)

Carnot

X(3652)

Suppose that A'B'C' is the cevian triangle of a point P. Then
LA, LB, LC concur if and only if P
lies on the cubic K455.

Suppose that A'B'C' is the anticevian triangle of a point P. Then
LA, LB, LC concur if and only if P
lies on the cubic given by the barycentric equation

Quim Castellsaguer defines the Carnot triangle as the triangle formed by the circumcenters of the triangles BCH, CAH, ABH, where H denotes the orthocenter. The Carnot triangle is also known as the Johnson triangle.

X(3657) = AYME PERSPECTOR

Let A'B'C' be the medial triangle of the reference triangle ABC. Let
P be the point of intersection of the lines X(1)X(3) and BC, and let P'
be the point where the line through P perpendicular to line AX(3) meets
that line. Let LA be the line PP', and define LB
and LC cyclically. Let A''=LB∩LC,
and define B'' and C'' cyclically. The lines AA'', BB'', CC'' concur in
X(3657). (Jean-Louis Ayme, Hyacinthos #16676, August 21, 2008)

The Ayme triangle A''B''C'' is also perspective to these triangles:
tangential, orthic, intangents, extangents, and the circumorthic.
(Peter Moses, November 7, 2011)

X(3659) = FEUERBACH POINT OF EXCENTRAL TRIANGLE

Let I be the incenter and IA the A-excenter of the
triangle ABC. Let LA be the line joining the circumcenter of
triangle BCI and the incenter of triangle BCIA, and define
LB and LC cyclically. The lines LA,
LB, LC concur in X(3659). (Seiichi Kirikami,
April 12, 2010)

Let Ea be the ellipse with B and C as foci and passing through X(1), and define Eb and Ec cyclically. Let La be the line tangent to Ea at X(1), and define Lb and Lc cyclically. Let A' = La∩BC, B' = Lb∩CA, C' = Lc∩AB. Then A', B', C' are collinear, and the line A'B'C' meets the line at infinity at the isogonal conjugate of X(3659). (Randy Hutson, April 9, 2016)

X(3659) lies on the circumcircle and these lines: 3,164
55,258 106,1130 759,1128

Let CA be the circumcircle analog of the Conway circle; that is, the circle with center X(3) and radius (R2 + s2)1/2. Then X(3660) is the radical trace of CA and the Conway circle. (Randy Hutson, October 13, 2015)

The discussion of combos near the beginning
of ETC is continued here. Suppose that T is a central triangle, and let
nT its normalization, so that the triangle nT is essentially a 3x3
matrix with row sums equal to 1, and the rows of nT are normalized
barycentrics for the A-, B-, C- vertices of T.

Let X be a triangle center, given by barycentrics x : y : z, not
necessarily normalized. The point whose rows are the matrix product
X*(nT) is then a triangle center, denoted by Xcom(T).

Among central triangles T are cevian and anticevian triangles and
others described at MathWorld. A brief list follows, with A-vertices
given in barycentrics (not normalized):

X(3720) = X(756)com(INCENTRAL TRIANGLE)

Barycentrics a(ab + ac + 2bc) : :

Let A'B'C' be the incentral triangle. Let A" be the intersection, other than the midpoint of BC, of the A-median and the bicevian ellipse of X(1) and X(2). Define B" and C" cyclically. The lines A'A", B'B", C'C" concur in X(3720). (Randy Hutson, December 26, 2015)

Continuing the discussion of points Xcom(T),
suppose that T is an arbitrary triangle, and let nT denote the
normalization of T. Let NT denote the set of these triangles, as
matrices, and let * denote matrix multiplication. Then NT is closed
under *. Also, NT is closed under matrix inversion. Consequently, (NT,
*) is a group, comparable to the group of stochastic matrices.

Suppose that T1 and T2 are triangles. In many
cases, the product T1*T2 is well-defined (e.g.,
TCCT, page 175). However, n(T1*T2) may not be
n(T1)*n(T2) if T1 and T2
are not normalized. Therefore, it is important, when dealing with
products, to include the "n" if it is intended.

As a class of examples of triangles defined by matrix products,
suppose that T1 is the cevian triangle of a triangle center
f : g : h (barycentrics) and that T2 is the cevian triangle
of a triangle center u : v : w. The A-vertex of the triangle
T3 = (nT1)*(nT2) is given by

u(gu + hu + gv +hw) : hv(u + w) : gw(u + v),

from which it can be shown that T3 is perspective to the
triangle ABC, with perspector

Thus, T3 can be constructed directly from the pairs {P,
ABC} and {Q, T2}. Regarding the possibility that
T3 is also perspective to T1, the concurrence
determinant for this condition factors as
F1F2F3, where

Having considered the group NT of normalized triangles, we turn next
to 3-point combos based on product triangles and inverse triangles.
Recall that such a combo, denoted by Xcom(T), is defined, as in the
preamble to X(3663), by the matrix product
(x y z)*(nT), where nT is
the normalization of T.

X(3751) = X(1721)com[INVERSE(n(HEXYL TRIANGLE))]

Let A'B'C' denote the inverse of the normalized hexyl triangle. Then
A', expressed in homogeneous barycentrics (i.e., not normalized), is
given by

2a(b + c) : a2 + b2 - c2 :
a2 - b2 + c2

In general, if a triangle DEF is perspective to a given triangle GHI
from a point P and also perspector to another triangle JKL from a point
Q, then, clearly, lines joining P to G, H, I, and line joining Q to
J,K,L concur in pairs to form DEF. Thus, if constructions are known for
G,H,I,P,J,K,I,Q, then a construction for DEF easily follows. In
particular, the triangle A'B'C' can be constructed in several ways, as
it is perspective to the following triangles with perspectors:

Many triangle centers can be defined (as just
above) by the form Xcom(nT), where T denotes a central triangle. In
order to present such centers, it is helpful to introduce the notation
T(f(a,b,c), g(b,c,a)) for central triangles. Following TCCT, pages
53-54, suppose that each of f(a,b,c) and g(a,b,c) is a center-function
or the zero function, and that one of these three conditions holds:

the degree of homogeneity of g equals that of f;
f is the zero function and g is not the zero function;
g is the zero function and f is not the zero function.

There are two cases to be considered: If g(a,b,c)=g(a,c,b), then the
central triangle T(f(a,b,c), g(b,c,a)) is defined by the following 3x3
matrix (whose rows give homogeneous coordinates for the A-, B-, C-
vertices, respectively):

The vertices of the five Euler triangles lie on the nine-point circle. (See C. Kimberling, "Twenty-one points on the nine-point circle," Mathematical Gazette 92 (2008) 29-38.) Randy Hutson observes (January 8, 2015) that the 2nd Euler triangle is the complement of the circumorthic triangle, the 2nd Euler triangle is the reflection of the orthic triangle in X(5); the 3rd Euler triangle is the complement of the 1st circumperp triangle, the 4th Euler triangle is the complement of the 2nd circumperp triangle; the 4th Euler triangle the triangle whose sidelines are the radical axes of each of the Odehnal tritangent circles (defined at X(6176)) and the corresponding excircle; the 5th Euler triangle is the complement of the circummedial triangle, and the vertices of the 5th Euler triangle are the intersections, other than midpoints of the sides of ABC, of the nine-point circle and the medians.

The notation T(f(a,b,c),g(b,c,a)) can be used to define several more
triangles; in each case, two of the perspectivities can be used to
construct the triangle.

Peter Moses noted (12/23/2011) that T(bc, b2) is
triply perspective to ABC. With T(bc, b2)
order-labeled as A'B'C', the three perspectivities with perspectors are
as follows:

AA'∩BB'∩CC' = X(6); AB'∩BC'∩CA' = P(6);
AC'∩BA'∩CB' = U(6). (The notation P(k) and U(k) refers to a
bicentric pair of points; see Tables at the top of ETC.)
Likewise, T(bc, c2) is triply perspective to ABC, with
perspectors: X(76), P(10), and U(10).

X(3817) = X(2)com(3rd EULER TRIANGLE)

Barycentrics 3b3 + 3c3 - a2b -
a2c - 2ab2 - 2ac2 - 3b2c -
3bc2 + 4abc

In the plane of a triangle ABC, let I = X(1) and
L = line through midpoint of CA perpendicular to BI
L' = line through midpoint of AB perpendicular to CI
L'' = line through midpoint of AI perpendicular to BC
The lines L, L', L'' concur in a point, A'; define B' and C' cyclically. Ten X(3817) = X(2)-of-A'B'C'.
Also, A'B'C' = complement of the excentral triangle, and A'B'C' = extraversion triangle of X(10). (Randy Hutson, September 14, 2016)

Let A'B'C' be the intouch triangle, and AaAbAc, BaBbBc, CaCbCc the A-, B-, and C-extouch triangles. Let Ab' = B'C'∩BcBa and Ac' = B'C'∩CaCb. Define Bc' and Ba', Ca', Cb' cyclically. Then X(3817) is the centroid of {Ab',Ac',Bc',Ba',Ca',Cb'} is X(3817). The six points lie on a common ellipse. (Randy Hutson, September 14, 2016)

Let Wa be the inverter of the B- and C-excircles, and define Wb and Wc cyclically; see X(5577). Let Ia be Wa-inverse of the incircle, and define Ib and Ic cyclically. The radical center of circles Ia, Ib, Ic is X(3817). (Randy Hutson, September 14, 2016)

Let A'B'C' be the circumcevian triangle of X(1). Let L1 be the Simson line of A', and define L2 and L3 cyclically. Let A'' = L2∩L3, and define B'' and C'' cyclically. Then X(3817) = X(2)-of-A''B''C''. (Angel Montesdeoca, June 28, 1017)

X(3822) = X(2099)com(3rd EULER TRIANGLE)

Barycentrics (b + c)(b3 + c3 - a2b - a2c - b2c - bc2 - abc) : :

Suppose that ABC is an acute triangle. Let LA be the circle that is externally tangent to the nine-point circle and to the sidelines AB and AC. Define LB and LC cyclically. The three circles are here named the Odehnal tritangent circles. Their radical center is X(3822). The center of the Apollonian circle of LA, LB, LC is X(6167). (Boris Odehnal, "A Triad of Tritangent Circles, Journal for Geometry and Graphics 18 (2014) no. 1, 61-71)

X(3861) = X(546)com(EULER
TRIANGLE)

As a point on the Euler line, X(3861) has Shinagawa coefficients (1, 13)

X(3861) lies on this line: {2, 3}

Inverse Triangles and
More Combos

The Euler triangle and the 2nd, 3rd, 4th, and
5th Euler triangles are discussed in the preamble to X(3758), along
with eight other central triangles using the notation T(f(a,b,c),
g(b,c,a)). Recall that the inverse of a normalized central triangle T,
denoted by Inverse(nT) or Inverse(n(T)), is also a normalized central
triangle. Barycentrics for the inverse of each of the 13 triangles and
properties of these triangles are given (Peter Moses, December 2011) as
follows:

For present purposes, label these last triangles as IT1, IT2, IT3,
IT4, IT5, IT6, IT7, IT8; then all except IT5 and IT6 are perspective to
the reference triangle ABC; perspectors are described at
X(3862)-X(3866). Label the corresponding original triangles T1, T2, T3,
T4, T5, T6, T7, T8; then the following pairs are perspective: (T1,
IT1), (T2, IT2), (T3,IT3), (T4, IT4), (T7, IT7), (T8,IT8); coordinates
for the perspectors are long and omitted.

The triangles IT1 and IT2 are triply perspective to ABC (as are T1
and T3). Order-label IT1 as A'B'C'. The three perspectivities are then
given by

X(3913) = X(10)com[INVERSE(n(INTOUCH
TRIANGLE))]

Let OA be the circle tangent to side BC at its midpoint
and to the circumcircle on the same side of BC as A. Define
OB and OC cyclically. X(3913) is the radical
center of the circles OA, OB, OC. See
the reference at X(1001).

As a point on the Euler line, X(4240) has Shinagawa coefficients ((E - 8F)F,6(E + F)F - 2S2).

X(4240) is the point of intersection of the Euler lines of twelve triangles, constructed as in the next few sentences.

Let E be the Euler line of a triangle ABC. Let A1 =
E∩BC, and define B1 and C1 cyclically. Let
AB be the reflection of A in B1, and define
BC and CA cyclically. Let AC be the
reflection of C in B1, and define BA and
CB cyclically. The Euler lines of the four triangles ABC,
AABAC, BBCBA,
CCACB concur in X(4240). (Dao Thanh Oai, Problem
1 in attachment to ADGEOM #1709, September 15, 2014). See also Telv
Cohl, 'Dao's Theorem on the Concurrence of Three Euler Lines,'
International Journal of Geometry 3 (2014) 70-73: Dao's Theorem.

Continuing, let A*B*C* be the paralogic triangle of ABC whose
perspectrix is E. Then X(4240) lies on the Euler line of A*B*C*. (Dao
Thanh Oai, noted just after Figure 1 in attachment to ADGEOM #1709,
September 15, 2014).

Continuing, redefine AB as the point on line AC and
AC as the point on line AB such that B1,
A1, AB, AC line on a circle and
A1, AB, AC are collinear. Define
BC and BA cyclically, and define CA
and CB cyclically. Let A2 =
BABC∩CACB and define
B2 and C2 cyclically. The Euler lines of the five
triangles ABC, A2B2C2,
AABAC, BBCBA,
CCACB concur in X(4240). (Dao Thanh Oai, Problem
2 in attachment to ADGEOM #1709, September 15, 2014).